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How would an Inverted Delta Wing planform behave, Straight part forward? Are there wings shaped this way?
There is info about straight edge wings, swept back wings, swept forward wings, delta wings.
Is there also info about a Delta wing with a straight leading edge and a tapered trailing edge, the point rearwards, like a manta ray?

To answer your second question first: Yes, these are mainly used for ground-effect craft. The idea is to have positive angle of attack, a tapered wing and a trailing edge with constant, low distance from the ground. This question covers their sweep angle.

The RFB X-114 ground-effect craft. Forward is right.
The wing would behave just like an ordinary, forward swept, low aspect ratio wing. It would have a small lift curve slope and a lousy stall behavior, because the low tip chord will make the tips stall first, causing the whole wing to roll. Unfortunately, the low leading edge sweep will prevent vortex lift, which is a typical feature of regular delta wings, to develop, so the stall angle will be only little higher than that of a straight wing with the same aspect ratio.

Manta rays are more diamond-shaped, so their "wing" has positive leading edge sweep and almost zero quarter-chord sweep. Their stall behavior would also be nasty if they were rigid, but the wings are actually fins and used for propulsion, not lift, so stall does not mean that the fish plunges to the ground. The picture above shows an Atlantic manta and was shamelessly copied from Wikipedia.

If the manta ray wing stalls, he plunges... millimeters deeper into the water. ;)

Wow, it took me a few minutes to figure out what I was looking at and from what angle, re the first picture.

Note that manta rays (or any other rays) don't have swim bladders so they do rely on hydrodynamic lift for depth control. But they only need a little bit of it and with dynamic pressure in water they don't need much speed to get it.

The X-114 is one of my favorite designs, for some reason.

Responses are good and detailed, but the Arup, Hoffman, Milt's Little Bird and Flying Pancake tailess airplanes all had this type of straight leading edge, and something like an inverted delta trailing side of wing, but were not ground effect machines as the Lippisch aerodynes. Any specific experiences or info about these low Reynolds number, low aspect ratio aircraft? Regards, Salut +

See also the '30s Arup planes, from Indiana. Straight LE, and semi-circular TE. A low aspect-ratio lifting body that did not suffer from high parasitic drag due to wing-tip vortexes, and were slippery and quick with very low landing speeds. Hatfield in the '80s followed them with his "little Bird" series which flew much the same.
 The '40s Eschelman diamond shaped "flying flounder" was another

The reverse delta wing configuration, which Dr. A. Lippisch proved to be very stable in ground effect, creates a very steady position of the Cm (pitching moment) when the wing is close to a surface.

I just realized the Peter Girard Delta shaped RotorWing F-104 VTOL considered at Ryan, could have also be flown with the rotor having an straight leading edge, but it seems never went beyond mockup stage. Thanks. Salut +

This report, silent, German text, about development of Alexander Lippisch X-113 Ground Effect machine, also able to free flying, answers a lot of question. https://youtu.be/gnflnBF7jWk Patent US3190582, 1965, describes this Aerofoil Boat. Blessings +

https://link.springer.com/chapter/10.1007/978-3-319-02836-1_7

References, from an article by Stuart Urquhart, and other by Afaq Altaf: -Lippisch, A. M., "Ground Effects Utilizing and Transition Aircraft," U.S. Patent US3190582A, filed 1965; Gerhardt, H. A., "Supersonic Natural Laminar Flow Wing," Patent US005538201A, filed 1996; Mou, H., "An Experimental Investigation of the Passive Control of Reverse Delta Wing Vortex Flow Structure and Aerodynamic Characteristics," Master's Thesis, McGill Univ., Montreal, 2015; Ko, L. S., "An Experimental Investigation of the Aerodynamics and Vortex Flowfield of a Reverse Delta Wing," Ph.D. McGill Uni, Montreal, 2016.

'The trailing vortices generated by a reverse delta wing with different wing configurations', T Lee, SM He. Aerospace Science and Technology. 2018; -'Study of the reverse delta wing', Afaq Altaf et al, Journal of Aircraft, Jan 2011; 'Aircraft', RJ Thompson, Patents US2167143;  US2303713. Blessings +

Reserarch indicates Lift to Drag ratio in a Reverse Delta Wing Planform is roughly same as in an ordinary Delta

The YouTube video of A Lippisch 'Aerodyne', patent CA574955, not that further developed by Collins and Dornier https://youtu.be/3Nk64aTEvAg?si=sXlQXKJL8veaU8Tt is amazing

https://youtu.be/gnflnBF7jWk?si=cc0gMULqPe-6qyrd   https://youtu.be/LMZIriem-xw?si=WFpIf1vAtRW-UPVn  Probably, with wheels embedded in wing floats and another in rear of fuselage, you could have an anphibious aircraft. Blessings +

: Besides book on Aerodynamics of Alexander Lippisch unconventional machines, see A Lippisch Wikipedia article, and patent US3190582, attached please find images of an Iranian Sea Patrol WIG and improvements in basic A Lippisch X-112 design to reduce drag [ Blessings +

From the plan-view's (less from the photo), it looks like they were taking inspiration in the design from the Klingon warships of Star Trek!

No 'Star Wars' machine, it's an Alexander Lippisch design, X-112 and X-113, patent US3190582, see videos: https://youtu.be/LMZIriem-xw?si=BMqoSnREf3D3Tnt8 https://youtu.be/gnflnBF7jWk?si=cc0gMULqPe-6qyrd Blessings +


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Modelling airplane wing
In the X-Z plane I have defined an airfoil shape using the function z = f(x). See figure 1. However, since the chord length of real airplane wings increases along the wingspan, I thought I needed to improve this function and found the following formula to describe the relationship between the tip chord, root chord, and wingspan:
$$x=\frac{\left(c_{root}-c_{tip}\right)}{\frac{b}{2}}y+c_{tip}$$
where $$c_{tip}=1\ \ \ c_{root}=3\ \ \ \ \ \ \ \frac{b}{2}=20$$
(The formulas shown are not important in figure 2. I just wanted to show the function f(x/(1+y/10)) but I can provide it to you if you want)

However, I first tried replacing x with 1+y/10 in my z = f(x) function, then I tried multiplying x with 1+y/10, but neither provided me with an accurate model. Later, instead of multiplying with x, I wrote it as $$f\left(\frac{x}{1+y/10}\right)$$ and it gave me figure 2. What I actually want to understand is the mathematical basis behind this phenomenon, that is, how could I express this situation in mathematical terms if I needed to?
I still don't know if this model is correct, because in figure 2 it seems to me that only the chord length has increased, but shouldn't the area of the airfoil increase? For example, when c = 1, Area = 2, and when c = 3, shouldn't Area = 6?

Is $b/2$ the half wingspan?  How is area calculated, and why does that even matter?

@CamilleGoudeseune Wings that are tapered are called tapered -- wings that swept are called swept.
Wings can be swept about the LE, the c/4, c/2, TE, or any other reference line.  If the reference line has zero sweep -- then it is considered un-swept.  Even if this results in a LE and/or TE that is swept.
Lots of low speed aircraft use tapered wings that are considered un-swept.  They often use the reference line as the location for the main (or aft) spar.  This allows primary structural members to be simple and straight, while still allowing for the aerodynamic benefits of taper.

@RobMcDonald, doh, of course.  No idea what I was thinking.

Is your goal to generate 3D points that lie on the surface of a wing, or to have a closed form analytical equation for a wing surface?
The first is perhaps a little easier.  When using this approach, I generate a "block" of points that I then apply transformations to.  Think of the wing like a chunk of clay -- start with a block, then stretch it, twist it, shear it, etc. until it is in the shape of the wing.
The points are initialized in a simple array (creating the block) and then the transformations are applied in sequence.
To generate an analytical form in geometry, we usually use parametric functions instead of explicit (or implicit).
Explicit functions are ones like $y=f(x)$.  If you wanted to define a circle explicitly, you would write $y=\sqrt{r^2-x^2}$ and $y=-\sqrt{r^2-x^2}$.  We need two parts because a circle fails the vertical line test when written as $y=f(x)$.
Implicit functions are ones like $f(x,y)-C=0$.  To define a circle implicitly, you would write $x^2+y^2-r^2=0$.  This is sometimes easy to write and work with, but a pain to actually evaluate the points on the circle -- you must perform an elaborate solution procedure for every point.
Parametric functions are ones like $x(t),y(t)$.  To define a circle parametrically, you would write $x=r\cos(\theta),\ y=r\sin(\theta)$.
If you wanted to stretch your circle, you could just apply a transformation
$x'=r\cos(\theta)$
$x=2\,x'$
$y=r\sin(\theta)$
Here, I use $x'$ to denote "x-prime", an alternate variable similar to x.  At other times in life, you will see a prime used to denote a derivative -- I'm not doing that here.
I think that starting with airfoil equations written parametrically (even if you substitute $t=x$ to have $z(t)$ and $x(t)$) will make your life easier.  You can then think about each of the transformations you want to apply as a single operation that can be built up one on top of another.
What software are you using to do this in?  It looks like Mathematica or MathCad or something.  I can help with the Math and Aerospace parts of this, but I don't have access to that software, so I can't try it myself.

I used Desmos 3D

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What type of plane/model is this twin-engine plane?
I would like to know what type/model is this plane, because I like it a lot.

That’s a photo out of Microsoft Flight Simulator 2020 of their default Beechcraft King Air 350i.
This version of the King Air is no longer produced by Textron Aviation, but it’s very similar to the current production iteration, the King Air 360/360ER.

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Why does thrust-to-weight ratio depend from max Mach number?
Why, for jet airplane design, does maximum Mach (or maximum speed) affect thrust-to-weight ratio so strictly? (as you can see for example in Daniel P. Raymer "Aircraft Design: A conceptual Approach" Table 5.3 page 80, where you can find $\frac{T}{W_o}=a\cdot M_{max}^C$, where “$a$” and “$C$” are empirical factors, for jet transport $a=0.267$ and $C=0.363$)
Just now I did notice to detail that Raymer say : "The equations were developed by the author using data from Ref 1" Ref1: i.e. the famous annual book : "Jane's All the world Aircrafts" . But anyway "maximum speed" (or maximum Mach number) in this case the maximum speed is ...the speed that the aircraft is not able to exceed  or.... the speed that is forbidden for the pilot to try to exceed because it can be dangerous? Because, if it's the first case it's more understandable; in fact, if the other parameters are fixed, with the same fixed distance, if T/W is low the acceleration is lower and so before the end of the distance, you can arrive to low speed. At the opposite if T/W is high the acceleration is higher and so before the end of the distance, you can arrive to high speed. What do you think? So, in Janes book, what is the definition of Maximum Mach number?

Sounds backwards...  T:W would drive the max achievable speed.  Even that assumes, though, that your Max achievable speed isn't limited by an MMO or VMO.  An F-15 probably doesn't care; many airliners can reach MMO in level flight even at max weight.

@RalphJ it just depends on which way you look at it. When designing an aircraft, max Mach would be a typical requirement, and the given equation will allow you to estimate the T:W ratio needed to achieve it.

but, why does each affect other?

How can we demonstrate the dependence each other (theoretically, with formula not only with empirical factors)?

Disclaimer: I have no copy of Raymer's book, so I have to guess what this table looks like.
There should be no direct dependency, except that all aircraft need to operate in the real world. If you look at the wing loading of regular jet transports, you will notice that they bunch together around 500 - 600 kg/m². This is the maximum wing loading which still allows them to operate from normal airports - go to higher values an you need a dry lakebed for take-offs and especially landings.
To push the resulting wing to higher flight speeds needs more thrust - hence the dependency. If you start with a higher wing loading (say, 800 kg/m²), the thrust/weight ratio for reaching the same Mach number is lower. The approach and landing speeds grow with wing loading as well, which in turn affects the resulting landing distance. I guess there is a maximum landing distance requirement for all aircraft in the table which is not explicitly mentioned.

Peter you say: " If you start with a higher wing loading (say, 800 kg/m²), the thrust/weight ratio for reaching the same Mach number is lower. " Is not the opposite?

@d.pensopositivo: No, the smaller wing creates less drag and needs less thrust. Once the aircraft flies fast enough, a high wing loading will make it more efficient.

@d.pensopositivo Peter Kämpf is correct.  At high speeds parasitic drag dominates, and this is proportional to the wing area.  If thrust and weight are held constant, higher wing loading -> less wing area -> less drag -> higher speed.  See the formula and derivation in my answer for exact details.

Related: https://engineering.stackexchange.com/questions/8101/how-can-we-prove-that-thrust-to-weight-ratio-depends-by-max-speed
In a short answer: In practice Mmax and Thrust to weight ratio are not correlated because other factors are more important, but in theory there is some dependency.  The answer depends if you're trying to compare planes, in which case you're looking at correlations among dozens of performance parameters and trying to fit estimates into simple equations.  Or you could be designing a plane, in which case you'd run equations to make sure the plane has enough thrust to meet the target Mmo. The case presented by Raymer seems to be a mix of the two.  He's using industry data to predict necessary T/W early in the design process.
Design Theory
The most confusing part here is that normally we think of thrust working against drag to provide acceleration, yet here we're talking about steady-state speed, not acceleration. The catch here is that drag is not constant as speed increases.  As you accelerate you reach a point where drag becomes strong enough to counteract the maximum engine thrust, and you can't go any faster in level flight (see the first graph [source])).

Assuming thrust is constant with respect to velocity, then from the formulas for drag as a function of speed we can derive  the maximum velocity.  This maximum velocity (in a thrust-limited level flight) can be written as a function of thrust-to-weight ratio and wing loading: (see derivation here (pdf), compare with here (pdf))
$$
V_{max}=\sqrt{\frac{\frac{T_{A,max}}{W}\frac{W}{S}+\frac{W}{S}\sqrt{(\frac{T_{A,max}}{W})^2-4C_{D_{0}}K}}{\rho _{\infty }C_{D_{0}}}}
$$
Where $ T_a$ is available thrust, $W$ is the weight of the aircraft, $\frac{W}{S}$ is wing loading, $K$ is the correction factor for induced drag, and $C_{D_0}$ is the drag coefficient of aircraft when lift equals zero.
However, you'll notice that there are other parameters in this formula besides thrust to weight ratio like wing loading and drag coefficient.  In practice these are not trivial considerations, especially when comparing different planes.  Raymer himself notes "Later in the design process, aerodynamics calculation of drag at the design maximum speeds will be used, with other criteria, to establish the required T/W."  There are also factors this formula doesn't consider like the variation in thrust as speed changes. These are the reasons thrust to weight ratio and maximum speed are not correlated when comparing different planes.
Comparing Planes
Let's see if this relationship actually holds out when comparing different planes. What real-world perfomance parameter should we reference for Vmax? Mmo, high speed cruise TAS, or something else? Technically speaking, Mmo is the certified max operational speed and is a safety limit, not just a performance limit.  During descents you can pick up energy very quickly and exceed the maximum level flight speed and Mmo. However, in practice (for business jets at least) the Mmo, high speed cruise TAS, and maximum level flight speeds are correlated. After all, you don't design many planes for their diving performance and many jets are designed to cruise close to their Mmo.
The truth is that none of these speeds is closely correlated with thrust/weight when comparing jets.  In my data set of 31 business jets from "Business Jets 2012", the correlation between thrust to weight and Mmo has a pitiful R-squared value of 0.20 (see the second graph). I'm not sure what data Raymer was using and if anyone has better data for showing a trend let me know.  I should note that once we account for the effects of wing-loading, there is a weak relationship between thrust to weight ratio and maximum speed.  So the effect of thrust/weight on Mmax in the above equation is very weak but present (see the third graph).

There are a lot of parameters that loosely correlate with Mmo besides thrust, including aircraft type, cruise altitude, wing loading, and range.  These parameters need to be engineered in conjunction to make sure the airplane meets all the performance expectations.  Mmo may be reduced or increased so that one of these parameters reaches its target and vice versa.
In theory, there is a correlation, but I have to say my stats don't agree with Raymer's on the variables being "closely related" when comparing different jets.  Other variables like drag and wing loading are too important to be neglected.

Talking about "maximum Mach number" I didn't understand if Raymer means Mach number coming from V(Emax) (for levelling unaccelerate flight) or someone among limit speeds. I suppose, because it recalls Referes from Jane's, that we can find in Jane's, but till now....

@d.pensopositivo I believe Raymer is referring to the maximum speed you want the engines to be capable of achieving in level flight during the design process.  I'm not expert on  how to choose that number and what it is in relation to Mmo, but I do know that for existing planes these top speeds and high speed cruises tend to be pretty closely related.

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About climb: If we use T/W as input, the second grade equation yields two solutions for W/S. Are both solutions realistic?
Raymer tells (Source: Raymer: Aircraft Design – A Conceptual Approach Chapter "5 THRUST-TO-WEIGHT RATIO AND WING LOADING" Paragraph “5.3 WING LOADING” Subparagraph “Climb and Glide”) that, when the aircraft is climbing, if the climb angle γ is small,
then $G=\frac{T-D}{W}$
(G is Climb Gradient as defined by Raymer words: "Rate of climb is vertical velocity,.. Climb gradient G is the ratio between vertical and horizontal distance traveled. This is approximately equal to the vertical climb rate divided by the aircraft velocity, or the sine of the climb angle γ." Source: Raymer: Aircraft Design – A Conceptual Approach Chapter "17 PERFORMANCE AND FLIGHT MECHANICS" Paragraph “17.3 STEADY CLIMBING AND DESCENDING FLIGHT” Subparagraph  “Climb Equations of Motion”)
Then $\frac{D}{W}=\frac{T}{W}-G$   ( that is the first relation for D/W).
At the same time  $D=q\cdot S\cdot c_{D0} + q\cdot S\cdot\frac{c_L^2}{π\cdot AR \cdot e}$
And substituting $c_L=\frac{W}{q\cdot S}$,
$$\frac{D}{W}=\frac{q\cdot S\cdot c_{D0}+q\cdot S\cdot\frac{c_L^2}{π\cdot AR\cdot e}}{W}=\frac{q\cdot c_{D0}}{\frac{W}{S}}+\frac{W}{S}\cdot\frac{1}{q\cdot π\cdot AR\cdot e}$$
(that is the second relation for $\frac{D}{W}$)
Equating the first and the second relation and solving for wing loading,
we have:
$$\frac{W}{S}=\frac{\left(\frac{T}{W}-G\right)± \sqrt {\left(\frac{T}{W}-G\right)^2-\frac{4\cdot c_{D0}}{π\cdot AR\cdot e}}}{\frac{2}{q\cdot π\cdot AR\cdot e}}$$
MY FIRST QUESTION: In this equation are both, the bigger and the lower solution values for $\frac{W}{S}$ realistic for aircraft?
After,  Raymer notes that the term within the square root symbol in the equation above cannot fall below zero, so the following must be true regardless of the wing loading:
$$\frac{T}{W}≥G+2\cdot\sqrt{\frac{c_{D0}}{π\cdot AR\cdot e}}$$
MY SECOND QUESTION: Raymer doesn’t consider that there are two solutions sourcing from the condition that the term within the square root symbol cannot fall below zero, the other one is:   $$\frac{T}{W}≤G-2\cdot\sqrt{\frac{c_{D0}}{π\cdot AR\cdot e}}$$. Why shouldn’t be considered? Could the second solution represent something realistic?
I HAVE A SUPPLEMENTARY QUESTION:
Relating to my questions above, I have to report two comments Raymer writes immediately after the last equation and inequality.: "This equation says that no matter how "clean" your design is, the T/W must be greater than the desired climb gradient! ....... Another implication of this equation is that a very "clean" aircraft that cruise at a high speed despite a very low T/W will probably climb poorly. A 200-mph airplane that flies on 20 hp can't be expected to climb as well as an airplane that requires 200 hp to reach 200 mph (unless the latter weighs ten times as much)"
Could anyone explain me better the last sentences, in particular, in which case T/W is very low with high speed? If we have high speed doesn't T/W should be high too?

I converted your equations into MathJax to make the question more readable. Could you please check that I converted everything truthfully?

In inequality inside squared root "greek pi"times"AR"times"e" should be at denominator, "CD0" should be at numerator.     Anyway, I'm really grateful for your work of conversion to MathJax, because I tried to use it, but I need getting use to that.

Finally, I think I wrote correctely the formulae for my intention of question. Since now, if someone understood better my questions , can try to answer.

You should include your last comments into your question (and then remove the comments by clicking the (X) that appears when you hover the comment with the mouse). You may edit your question by using the editlink between the question tags and your signature block.

I saw "second grade equation" in the title and thought I'd be able to follow along. Sadly not... :(

I followed your suggestion, mins, I inserted additional question inside the test of main question.

Raymer's G is a little unusual. In effect, it is the tangent of the flight path angle $\gamma$, and since his equations are for small values of $\gamma$, we can assume that it is approximately equal to $sin\gamma$ or $\gamma$ itself when expressed in radians.
$$G \approx \gamma$$
Next, the expression $\pi\cdot AR\cdot e$: This is part of the square root of the lift coefficient for minimum drag. The complete equation is:
$$c_{L\:(minimum\:drag)} = \sqrt{c_{D0}\cdot\pi\cdot AR\cdot e} = c_{L\:(md)}$$
Next, I need to express that thrust-to-weight ratio differently. As long as we keep $\gamma$ small:
$$\gamma = \frac{T}{W} - \frac{c_D}{c_L} => \frac{T}{W}-\gamma = \frac{c_D}{c_L}$$
Now I can rewrite the equation of your first question:
$$\frac{W}{S} = q\cdot\frac{\frac{c_D}{c_L}\cdot \frac{c^2_{L\:(md)}}{c_{D0}}}{2} ± q\cdot\frac{\sqrt {\left(\frac{c_D}{c_L}\right)^2\cdot\left(\frac{c^2_{L\:(md)}}{c_{D0}}\right)^2 - 4\cdot c^2_{L\:(md)}}}{2}$$
and when I restrict the equation to the minimum drag point where $c_{D0} = c_{Di}$, I can write
$$\frac{W}{S} = q\cdot c_{L\:(md)} ± q\cdot\sqrt {c^2_{L\:(md)} - c^2_{L\:(md)}}$$
Fancy that! The expression under the root becomes zero when I look at the polar point of minimum drag! The only variable term under the square root is the ratio $\frac{c_D}{c_L}$, and it reaches its minimum at the $c_{L\:(md)}$ point. When I move away from this polar point, I should get two valid solutions.
I guess these two wing loadings allow to fly with the specified flight path angle at the given polar point and thrust-to-weight ratio. Since the weight is prescribed by the thrust-to-weight ratio, the solution is actually for two different wing areas. The smaller wing requires a higher speed and the larger wing a smaller speed than what you get at the lowest drag point. Between both, the speed and wing area should result in configurations which achieve a higher flight path angle $\gamma$, and outside of this interval no level flight should be achievable with the resulting wing areas and speeds.
I cannot follow how you arrive at the second question. The term under the square root should be
$$\left(\frac{T}{W}\right)^2 - 2\cdot\frac{T}{W}\cdot G + G^2 ≥ \frac{4\cdot c_{D0}}{\pi\cdot AR\cdot e}$$
But when I insert again as explained above, I can write for the term under the root:
$$\left(\frac{c_D}{c_L}\right)^2 - \frac{4\cdot c^2_{D0}}{c^2_{L\:(md)}}$$
so the condition for a positive root is
$$\frac{c_{D0}+c_{Di}}{c_L} ≥ \frac{2\cdot c_{D0}}{c_{L\:(md)}}$$
or
$$ \frac{c_{D0}}{c_L} + \frac{c_L}{\pi\cdot AR\cdot e} ≥ 2\cdot\sqrt{ \frac{c_{D0}}{\pi\cdot AR\cdot e}}$$
which is certainly true; both sides of the equation are equal at the lowest drag polar point and away from it the left side is bigger and the right side stays constant.
If you just isolate the term under the root and look at it by itself, I doubt you have something that corresponds to a realistic configuration. You need to extract both numerator and denominator to keep the physical analogies intact. What Raymer does is to look at the term under the root in isolation and then postulates that the first summand must be larger than the second. He puts both on the two sides of an inequality and extracts the roots on both sides separately. Then he moves G over to the other side. There is no second version of the inequality possible.

Peter Kämpf first of all, thank you for your answer, it was very helpful. When you say: "The ratio CD/CD reaches its maximum at the CL(md) point, so the first summand of the root will always be smaller than the second, except for the polar point at minimum drag when both are equal."  CONTINUE

CONTINUE What do you mean? First I suppose you mean "CD/CL" instead of "CD/CD". If it so, it seems to me that, at min drag CD/CL is min, not max, (CD is numerator) then, out of that point of minimum drag we have 2 solutions.

@d.pensopositivo: You are right, thanks for catching this. Corrected. If the root becomes zero, it does not matter if there is a + or a - ahead of it. Therefore I think there is only one solution. Or two with the same value, if you want to see it this way.

Peter, About the sentence "The ratio CL/CD reaches its maximum at the CL(md) point, so the first summand of the root will always be smaller than the second, except for the polar point at minimum drag when both are equal." I think: In the fisrt addendum under the root the variable part is CD/CL, so I think it should be : "so the first summand of the root will always be bigger than the second" instead of "smaller" . Do you agree?

Anyway as I showed above, Raymer considers the existence of three cases: 1) Two different solutions : when T/W≥G+2⋅(cD0/(π⋅AR⋅e))^(1/2) 2) Two coincident solutions : when T/W=G+2⋅(cD0/(π⋅AR⋅e))^(1/2) 3) No solutions : when T/W≤G+2⋅(cD0/(π⋅AR⋅e))^(1/2) . My objection in "my second question" remind that such condition is coming from a resolution of a second grade disequation i.e. (TW−G)^2−4⋅CD0/(π⋅AR⋅e)≥0, then T/W≤G-2⋅(cD0/(π⋅AR⋅e))^(1/2) is another part of condition coming from case 1) (unless there are other constrains not explicitly mentioned there).

@d.pensopositivo: Yes, I agree. I did not notice that it is really only the $\frac{c_D}{c_L}$ which changes in the root. Nevertheless, the root of a sum is not simply the sum of the roots of the summands - my equation for your second question looks different. I changed the answer accordingly.

I return on my second question. In order to have real solutions       $(\frac{T}{W}-G)^{2}-\frac{4Cdo}{\pi ARe}\geq 0$. Then for simplicity of reasoning I define $(\frac{T}{W}-G)=x$ and $\frac{4Cdo}{\pi ARe}=b$. The condition becomes $x^{2}-b\geq 0$, but this (second grade) inequation is verified in both following zones: $x\geq \sqrt{b}$ and also $x\leq - \sqrt{b}$ i.e. $\frac{T}{W}-G\geq 2\ast \ \sqrt{\frac{CD0}{\pi \ast AR\ast e}}$ and also $\frac{T}{W}-G\leq-2\ast \ \sqrt{\frac{CD0}{\pi \ast AR\ast e}}$. Raymer excludes the second one, I don’t know why.

authors: Василий Лебедев

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What resistance is offered by an appendage on an aircraft flying at 30,000ft at 200 mph?
What formula can I use for calculating the resistance offered by an appendage above an aircraft wing? The appendage has a surface area of 10 square meters, positioned at an angle of 45 degrees. The aircraft is traveling at 200 mph, at 30,000 ft.

200 mph IAS or TAS?

If your indicated airspeed is 200 mph use:

Drag = 1/2 × Area × coefficient of drag × V$^2$

If the true airspeed is 200 mph use

Drag = 1/2 × Density × Area × coefficient of drag × V$^2$

Plug the dimensions of your shape into the drag equation.
Coefficient of drag can be estimated based on literature, or can be derived through wind tunnel tests.


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How can I estimate thrust?
I am stuck on an aircraft performance question.
The question seems fairly simple, calculating the ground run between V_0 and V_rotational. I have attempted this a few ways but made the most progress using the average acceleration method. However, I'm still stuck on the fact that I am not provided with a thrust force.
The question outlines that the aircraft is electric, single engine, and propeller driven. Within the question the following values are also given: maximum take-off weight, wing span, wing area, engine power, drag coef, ground factor, and friction coef.
I'm not sure if I'm missing something but I've spent hours online going in circles about how I can get the thrust figure. As it seems to be the only thing I'm missing.
If anyone could nudge me in the right direction that would be great :)
Thanks

Can you show us your try? Anyway this answer should help.

You can calculate $T$ from $V_{TO}$, the shaft power $P_s$, and assuming a propeller efficiency $\eta _ p$, $V_{TO}$ is $1.2V_s$ and $V_s$ is easily calculated. See this example at the end of page 5 "This is a propeller driven aircraft with a shaft (brake) horsepower of 2400 per engine. We want to calculate the minimum ground run". Remember we can't help you if you don't show your own attempt explaining your detailed reasoning.

Is any take-off run data available for a similar aircraft?  Is this a model or full scale?


text:
Can anyone identify this biplane in 1940s Orange, MA?
The picture was taken at Orange Municipal Airport in Orange, MA, in the 1940's.  I can't identify the logo either.

I know it isn't, by my first impression of the logo on the vertical stabilizer was that of the Indianapolis Motor Speedway.

The emblem on the tail is a logo for the Argo airplane, however I can't find a good high resolution copy as they didn't use it in their advertising at all, and a different logo was used on their Blue Bird.
It is a pair of wings with ARGO written in an arc over it, and while the Blue Bird logo has a star in the middle, this doesn't - but it could be a ship, as the most common device associated with Argo is a Phoenician ship.
Their advertising really plots the decline of the company - full page ads with pictures while Alliance, small text-only ads while Hess, and nothing while Warrior.
Alliance's logo was a very simple overlapped double A.


text:
Why do cargo airplanes fly fewer cycles then airliners?
Reading Why do cargo airlines frequently choose passenger aircraft rather than aircraft designed specifically for cargo?, some answers say that cargo aircraft fly less than passengers aircraft.  Why?

It's due to the overnight parcel distribution model which limits the available operating time.  A parcel picked up by 5pm needs time to get a terminal, and then has to get to the destination distribution terminal in time to be distributed and dropped off by mid afternoon the next day.
This means the heavies in the freight business do their flying overnight running between their distribution hubs (being a "Freight Dog" pilot is considered a night shift sort of career with fewer hours flown per year than pax flying, maybe 300-500 hours vs 700-1000 flying pax) so that freight can arrive at distribution points in the morning where it goes out to sub terminals on smaller aircraft during the day, or sent out by truck.
For example, Fedex runs aircraft in and out of Atlanta only between 10pm and 6am.  For many years I worked at an airport with a Fedex Terminal and you would wonder how anything got done because all you ever saw during the day was parked MD-11s lined up in front of the terminal, apparently just sitting there, with no freight movements to speak of after about 7 am, except for smaller a/c flying sub-distribution runs through the morning.  Well they are actually just waiting for their outgoing freight to arrive and if you happened by at midnight, it was pretty busy and there was a fair number of a/c coming and going.
So a freighter has maybe an 8 hour operating day, whereas a passenger a/c gets flogged like a mule and starts flying at 6 am and runs until 9 pm.  Plus the loading and unloading process for pallets is somewhat slower than with self-propelled human cattle getting on and off for you.  So you can't really do more than a couple of legs per day.
To provide context, this UPS fact sheet shows a fleet of 255 UPS jets doing 318 daily trips on average
It says "In-Out" trips for that 318 number, so lets assume that's actually 2 legs, 1 in and 1 out, for 636 actual flight cycles.  That's only 2.4 cycles per day overall, almost as easy a life for the airplane as corporate.

This is fairly specific to US domestic operations. What about international routes (e.g. Cargolux)? They should operate all day, right?

Also is this why there are much fewer night flights than daytime flights?

Well, more like North American operations - US/Can/Mex.  I can't speak for Europe but I would think that internal cargo operations in the EU would follow a similar model. I am familiar with the conversion of old beaten up Regional Jets into cargo carriers for EU operations, and I remember it was the same kind of thing - low cycles and night flying.  The low cycle operations meant that high cycle airframes were marketable.  Oceanic operations are another thing entirely I would think.

@user3528438 you mean pax flights?

@JohnK Yes, passenger flights. I sleep well on airplanes so I prefer overnight trips more than daytime trips, but I always have hard time finding overnight tickets, e.g. from LA to Chicago.

A lot of major airports have curfews that ban operations between say 11pm and 6am so "red eye" flights have to be scheduled around those limitations.  Also since such a large block of pax, maybe a majority, are traveling on business, and most people don't sleep very well sitting in an airline seat (I certainly can't) it doesn't really work to take an overnight if you have to attend meetings next morning or afternoon with no sleep.  So I don't think the overnight demand is really there.

@JohnK I see. Thanks.

Unlike passengers, packages don't really mind sitting around in their "departure lounge" for hours (or even a day or two, if it's not overnight delivery) before getting on the plane.

@jamesqf OTOH, many packages show up early because there was space today to take some cargo sitting around that wasn't scheduled to go until tomorrow. Passengers don't generally work that way. That's why cargo can reliably fill one large plane per day despite uneven demand.

But what about all those packages whose senders didn't pay extra for overnight shipping?

They sit around longer

@John K: Or maybe not.  The shipper probably puts all the overnight parcels in the plane first, then fills up the remainder of the space.  So (as happens to me reasonably often) the thing I ordered with "regular shipping" shows up the next day.


text:
What's the airplane with the smallest ratio of wingspan to fuselage width?
I want to know (preferably within the General Aviation category) which airplane has the smallest wingspan compared to its fuselage width.
(I'm curious of the performance of such an airplane)

If we look beyond GA, the XF-85 goblin (and to a lesser extent other parasite fighters) comes to mind, though I suspect it’s not the smallest ratio produced.

Even though not an airplane in the full concept, the space shuttle is also a candidate.

The "flying bathtub" comes to my mind.  It was only a prototype though.  https://en.wikipedia.org/wiki/NASA_M2-F1

I'm curious what the BD-5/BD-5J comes to?

Do helicopters and rockets count? And what about a zeppelin?

I think the Stits SA-2A Sky Baby should be in the running.

wow, never heard of it, looks like a toy.

It's also in the running for prop bigger than wings!

Excluding the cowl bumps, it's about 3.2.

Omg that is so cute

A well-known GA airplane that looks like a blended wing body is the Dyke Delta homebuilt.  Measuring Wikipedia's three-view of the JD-2 variant yields a wingspan to fuselage-width ratio of 3.0 to 3.7, depending on how you argue about fairings.

The similar Wainfan Facetmobile is a lifting body, for an unbeatable ratio of 1.0.  (Conversely, a flying wing's ratio could be said to be infinity.)

Ahhh, yes.  The Facetmobile.  Looks like a paper airplane... Neat!

I was gonna post about lifting bodies but I think this covers that.

@ThomasPerry the Facetmobile's origami, never mind a lack of compound curves, has made it a popular subject for radio-control models.

which airplane has the smallest wingspan compared to its fuselage width?

By definition it would be an airplane where the fuselage is the wing. Such kind of aircraft where extensively flown during the '50s to understand the aerodynamic characteristics of the future re-entry vehicles and are termed lifting body aircrafts:

With a ratio of 1:1 they are formally unbeatable.

Anyway if we restrict our search to GA airplane then, according to the Guinness world record, the Bumblebee Starr II with a wingspan of 1.68m (5 ft 6 in) is the world's smallest piloted airplane. Unfortunately it crashed due to an engine problem and the picture you find around with the plane in yellow-black stripes is the Bumblebee Starr I which was a bit bigger. Anyway this should be it with Starr himself (source):

With a wingspan of 1.91m (6 ft 3 in) the Sits DS-1 Baby Bird is the world's smallest monoplane:

Missing official measures, I'd say that here we are around a ratio of 3:1 I'd guess.

Ratios?  Excluding cowl, or lack thereof.

Wow, is that prop hub bizarre because the two blades are separate?!  ... ah, EAA says "ground adjustable", presumably in pitch, thus separate blades.

Refreshing that you mentioned these examples.  The next question is obvious:  Will it kill you?  A sensible deterrent to the sane person... Excellent examples, tho...

@chux-ReinstateMonica, fuselage width.  I've edited the question's title to clarify.

@CamilleGoudeseune "smallest ratio of wingspan to fuselage width" --> OK, so a width-to-width ratio and not necessarily a small ship?  E.g. the space shuttle ratio of 78/18 or 4.3

thanks @sophit form bringing up the term "lifting body", I had heard it before but didn't recall it, and definitely it was appropriate to bring it up to this post.

And just for comparison, a military aircraft. Not general aviation, tho... Yes, this is the F-104.  Approx. 3.27 +/-.

On that theme, I was curious how the Phantom scored, being the Triumph Of Thrust Over Aerodynamics. Turns out its wings aren't quite as stubby as a Starfighter though, even though the body is fatter.

@Graham   Answer: - about 4 +/- ...

More than a few of them aren't military.  +1.

I came here to mention the rare GA version of the F1-4 (cough). This link suggests 2.5:1 https://www.britannica.com/technology/F-104.  16.7:6.68

@RussellMcMahon  Yep... they are mentioning fuselage length while the current consideration is fuselage width.  Hence the difference... Oh, and that GA version of the 104... No laxative needed!

You might consider the 1955 Pritchard Rocket Air Ship, (which was neither a rocket nor an airship) with the major caveat that it never actually got off the ground.

source

source

Charles Pritchard, designer builder and pilot, makes first test of his
wingless Rocket Air Ship on mile lone runway at Emporia, Virginia,
airport, 8A-1030A today, (Friday). Plane has short baffles or fins in
place of wings, 90 hp engine, conventional propeller and tail
assembly, tricycle landing gear, is 21 feet long 98 inches wide weighs
800 lbs. Pritchard made eight runs along runway, but didn’t; get off
the ground. Says he’ll make modifications and try again.

The CAA eventually made him stop trying.
Alternatively, for a plane that actually got airborne, there is the Model C Speedbird (N38C or NX38C) by Merle Larson, a former WW2 B-24 pilot who modified a Taylorcraft BC-12. The wingspan is reported as 12' while the fuselage is approximately 4'2" wide, or was before it crashed.

An All-American master craftsman, Merle Larson took things in his
hands when he decided to test a new wing  of his own design design in
1953. Larson took a mundane Taylorcraft BC-12 and replaced its standard wing with this very tiny thing. His wing was designed to work
within the high-speed slipstream of the propeller to generate more
than the normal lift. Then came the time of truth. The “plane” was
doing quite well on its maiden flight, but guess what?, yeah, the
engine suddenly stopped. The Speed Bird went down like a proverbial
brick and crashed trapping Mr. Larson, thankfully unhurt.
source

Yes indeedy, that modified Taylorcraft would glide hardly better than one whose stock wing had fallen off!

Maybe not a winner, but a not-to-be-missed aircraft nonetheless, the Gee Bee Model R Super Sportster

Image source
With a wing span of 7.62 m and a fuselage width of 1.98 m, it ends up with a ratio of 3.85.


text:
Control surfaces on radio-control models of airliners
Do large RC airliners like the huge Emirates RC A380 still need all the same control surfaces that a real A380 has I,e leading edge slats? If not why is this the case.
Many thanks for your time in advance
Lee

Consider an edit-- a leading-edge slat is arguably not a control surface.

Also the term you want is probably l.e. flaps not slats--

Related : https://aviation.stackexchange.com/questions/35723/why-do-model-aircraft-fly-and-maneuver-so-differently-from-real-aircraft

@quietflyer Slats (and other high-lift devices) are sometimes grouped under secondary control surfaces (e.g. SKYbrary says "Secondary flight controls are intended to improve the aircraft performance characteristics or to relieve excessive control loading, and consist of high lift devices such as slats and flaps as well as flight spoilers and trim systems.").

@Bianfable -- good point

High-lift devices (slats and flaps) are normally used also as control surfaces on fighters. On jetliners they are used mainly, if not exclusively, as... high-lift devices. Since we are dealing here with an A380 then I suppose it would be better not to call them control surfaces.

If you fill a scale-down model A380 with helium it could also fly without an engine.

I've seen several r.c. airliners first-hand and none had leading-edge flaps.  It's a matter of scaling.  Kinetic energy scales according to velocity squared and these r.c. planes don't land at the same velocity as their full-scale counterparts, so stopping distance is less of an issue and the need to slow down the touch-down speed a bit more vanishes.
Same reason that the r.c. versions usually also dispense with wheel brakes, thrust reversing (some actually do have this, the electric ducted fans are simply run in the reverse direction), elaborate Fowler flaps (simpler hinged flaps serve instead), etc.
There are also Reynold's number effects to consider, but we really don't need to get into that to understand why features such as operating leading-edge flaps are usually omitted from r.c. model airliners.
In any oddball cases where the pursuit for scale accuracy is carried so far as to include features such as leading-edge flaps, it is clearly not a matter of a need for such features!
Here's a related ASE link-- Why do model aircraft fly and maneuver so differently from real aircraft?

I've noticed that certain features in an RC model may not be required aerodynamically, but the designer/builder includes them for realism.  I'm amazed at the level of detail that some of these people achieve.

KE = 1/2 mv^2, but v is linear in m, so that's why KE varies with v^3 not v^2?

@CamilleGoudeseune -- I didn't realize until reading your comment that I had typed "cubed" instead of "squared", was simply an error--


text:
Can an F-35 perform a cobra maneuver using its lift fans?
Can an F-35 perform a cobra maneuver using its lift fans? I have no experience flying F-35s I fly a F-18 Super Hornet.

I would assume no, because the computers wouldn’t allow such angles in VTOL mode. Not sure, but my guess.

"I fly a F-18 Super Hornet." Indeed? Where are you based?  Surprised that you don't have buddies from pilot training who could provide an answer to this question for you faster than an internet forum.

Oceana Virginia is where I am, and unfortunately I don't know anyone who flies f-35s

I am also here to somewhat enlighten others but I also have questions

The F-35 however can perform 'Dojo Drift' as well as 'Pedal Turn' within its dynamic AOA limit, up to 50 deg. Your F-18 is capable of performing 'Pirouette' maneuver if stick and rudder is applied in the same direction at high AOA.

No.

The LiftFan is cleared for flight up to 250 knots.

The cobra ... must be entered from fairly high subsonic speeds.

So the fan may not be used to enter a cobra.  Once in the cobra, the very high AoA by itself is so amazingly draggy that a few more tens of thousands of pounds of thrust hardly matters.

Thank you I wanted an answer and this was very helpful.


text:
Can a Boeing 747 or A380 climb with only one engine?
Assume that a Boeing 747 or Airbus A380 takes off at max takeoff weight. Minutes after takeoff, 3 engines fail, leaving only 1 engine operating.
Can the aircraft still achieve a positive rate of climb?

For a very short period of time using momentum and existing airspeed, but I don't think either aircraft can climb on a single engine. Hopefully Terry can chime in on this one.

At max weight I doubt even 2 engines could produce a positive rate of climb.

In a takeoff configuration at or near MTOW, no; it would probably require at least two engine operative in these conditions plus low altitudes and air at or near STP.  After a few hours of fuel burn in a cruise configuration, most likely yes.  Lighter weights plus cleaned up would allow for a positive rate on one engine.

@RonBeyer I think you're correct, although I have no documentation on that condition. My guess is that for the 747, the problem would be dumping fuel fast enough to stay in the air.

I thought we could just check the thrust and weight, but it's not that easy. The static thrust of one engine looks like it is enough for both aircraft:

B744 has MTOW 4050 kN, estimating L/D to 18 it needs 225 kN to maintain altitude and the static thrust is 276–282 kN (depending on engine option).
A388 has MTOW 5640 kN, using the same L/D it needs 313 kN to maintain altitude and the static thrust is 332–356 kN (depending on engine option).

However, jet engine thrust changes with speed and altitude:

With altitude it first remains constant as long as the peak pressure is the limit and then starts to decrease with density.
With speed is first decreases, but then starts increasing again as it takes advantage of ram pressure recovery, which grows with the Mach number.

At low altitude where the engine still produces the rated thrust the best L/D speed will correspond to fairly low Mach number, around M0.3–M0.35, which is just above the minimum thrust point. IIRC the reduction is about 20% (exact profiles for specific engines don't seem to be published anywhere; occasionally some research papers have it for older engines, but not the new ones), and

282 × 0.8 = 225, so for B744 with the most powerful engine option it is just barely enough to maintain altitude.
365 × 0.8 = 285, so definitely not enough for A388.

Plus the L/D is probably worse then I estimated above due to the deflected ruder and ailerons needed to compensate for the asymmetric thrust. So, no, it's not enough (L/D 18 is typical value for clean airliner; exact numbers are hard to get).
At cruise altitude the Mach number would be around M0.7–M0.75, where the engines are above their static thrust already, but at those altitudes the engines are only producing about ¼ of the thrust, so you really need all of them. In fact the cruise altitude is always selected to be just below the operating ceiling, and operating ceiling is where the aircraft can no longer sustain more than 500 ft/min climb with all engines operating. If you lose even one engine in cruise, you always have to descend.

I doubt that either aircraft can climb on a single engine (though momentum and airspeed could help, as @RonBeyer already commented). The closest an aircraft came to this condition was the British Airways Flight 9 when it ingested volcanic Ash.
During this incident, the aircraft flew on one engine for around 90 seconds. In this case, a report from the flying magazine says:

On one engine they could maintain height at 13,000 ft; over the next five minutes, the next three engines relit, and they begun a climb.

This seems to indicate that the aircraft was not able to climb with one engine. However, during this incident, the aircraft was not climbing, but was descending at 1800m per minute.
Note: There seems to be some confusion about the altitude. The captain's website gives the altitude as 12,000 ft.

Actually, it was this incident which inspired my question.

No it cannot climb with only one engine.
If it can climb with one engine the plane would be a twin jet not a quad jet. The number of engines is determine by power require for normal operation plus a margin in case one engine malfunctions during climb out. An engine is a very expensive piece of equipment both in term of up front cost and maintenance so two engines are cheaper than four.
If the engine is powerful enough to allow climbing with only one engine the manufacturer would make it a twin, not a quad. Simple as that.

This is somewhat misleading. A twin jet still needs to be able to climb with at least a 2.4% climb gradient in the second segment climb and 1.2% in the fourth segment climb after an engine failure (see e.g. this answer). This question only asks for positive rate of climb.


text:
Where can I find pictures of "endplate"?
Where can it possible to find pictures or drawings of "endplate"? (for "winglet" is much easier, but not endplate)
I'm asking what's written above, after I read the following text from Raymer and I need a better comprehension of it.

The wing aspect ratio "$A$" is the geometric aspect ratio of the
  complete reference planform. the effective aspect ratio will be
  increased by wing endplates or winglets
Endplate: $A_{effective} = A(1+1.9\frac{h}{b})$ - (12.10)   where $h$ = endplate height
Winglet: $A_{effective} = \approx 1.2A$  -     (12.11)
These effective aspect ratios should also be used in the induced drag
  calculations below. note that Eq. (12.11) for winglets is a crude
  approximation based upon limited data for wings of moderate aspect
  ratio

......................

The extra drag due to non-elliptical lift distribution and the flow
  separation can be accounted for using e, the "Oswald span efficiency
  factor". This effectively reduces the aspect ratio, producing the
  following equation for $K$
$K=\frac{1}{πAe}$ (12.48)
If the wing has end-plates or winglets, the effective aspect ratio
  from Eq. (12.10) or (12.11) should be used in Eq. (12.48)

(Source: Daniel P. Raymer "Aircraft design . A conceptual approach" , Chapter 12 Aerodynamics)

The endplate that i think the OP is referring to is the the part that maintenance attaches to the wing to cover the end of the wing when a wingtip is removed, due to damage for example.

I seem to remember that wing end plates were sometimes used on Short Take Off and Landing(STOL) aircraft in the 1960's.
I found a few photos while doing an internet search but they seem to be much more prevalent on race car wings these days.

Here is an explanation of why end plates are used on race cars: Analysis: Rear Wing Endplates

The endplate has two functions, the primary reason for their shape is
  to control the wingtip vortices at the trailing edge of the wing as
  this reduces drag and also lift. The second reason is that they
  control how the airflow makes its way over, under or around the
  aerofoil device that is creating the downforce on the car, resulting
  in better airflow management and maximising the performance of the
  component.

Winglets and aerodynamic endplates are in the category of "wingtip devices". Try searching for "wingtip device" + fence or endplate


text:
Why does RPM increase when you lean the mixture in constant speed propeller engine?
In constant speed propeller aircraft like Diamond star DA-40, during run-up checks the AFM mentioned to increase throttle to increase RPM to 2000 and then do magneto check then rpm check then mixture check which to lean it and watch for RPM increase EGT and CHT increase .. although RPM increase by 20 to 50 RPM .. i am curious to know why

This isn't unique to constant speed props - it's a function of reciprocating engines, including those in cars etc.  See this answer: https://aviation.stackexchange.com/questions/73066/why-does-pulling-the-mixture-to-idle-cutoff-increase-the-rpm-slightly

Sorry, fat-fingered the review... meant to mark close as dupe. My vote shows "leave open", that's not what I intended.

It's because at the power setting you are using for those checks, the prop is still "off the governor"; the blades are still at their fine pitch stops, because the governor is calling for max RPM, say 2700 (you have the prop control full forward), and you only have enough throttle for 2000 even with the blades at full fine pitch.
So in effect, your propeller is still a "fixed" pitch propeller (sitting against the fine pitch stops trying to let the engine wind up to the governor setting, but there isn't enough torque present to get above 2000) and so RPM is sensitive to torque changes like with a regular fixed pitch prop.
When you add throttle for take off, RPM will increase as the blades stay on the fine pitch stops until you get to 2700 (or whatever the setting is), then stop rising above 2700 as you continue to add throttle.  At the point where RPM stops rising, you are "on the governor" and the blades are coming off the stops to regulate the load on the engine to keep the RPM at 2700.

I think OP wants to know why leaning at a constant throttle position results in a few RPMs increase.

I thought the question was about why there is an RPM increase at all (mixture, mags, whatever) when it's a CS prop. We need the OP to clarify.


text:
Why oil pressure decreases when coarsen the pitch during run-up?
During run-up checks, we set the
Throttle such a manner that RPM IS 2000 then we do run-up checks

Magneto check
2.prop check
Notice RPM drop
MAP increases
Oil pressure decreases

Why does MAP INCREASE AND OIL PRESSURE DECREASE ?

When you do the prop cycling check, pulling it to min RPM and having the prop make that cool "zzzhheewwww w  w  w  w  w w" sound, manifold pressure goes up because of the reduction in air velocity in the intake ducts as the big air pump that is the engine slows down.  Basically, the vacuum level goes down, which is an increase in manifold pressure closer to ambient.
The oil pressure decrease is from the oil flowing into the propeller to drive the blades toward low pitch.


text:
What is the difference between Jeppesen NavData & Digital Terminal Charts?
https://ww2.jeppesen.com/update-cycle-and-effective-dates-schedule/
in the link provided above,
Jeppesen NavData has 28day cycle whereas
Digital Terminal Charts have 14day cycle
Can you provide the reason for varying update cycles
i am aware by the name Terminal charts means airports information and approach plates, SIDs & STARs but what does NavData contain ?

Terminal charts are published per a 14-day cycle which is usual in IFR chart circulation.
Jeppesen states on their website that the NavData pack is published according to the 28-day AIRAC cycle.
The NavData pack contains navigational charts, terrain/obstacle data, and data for RNP/LPV.

I will respectfully disagree. The AIRAC cycle is 28 days, which is a worldwide standard. Jepp chooses to publish charts every 14 days, which allows them to update a chart instead of issuing a NOTAM. However, procedures are updated on the normal AIRAC cycle which includes both the chart and data.

Agree with Timbo. That's an option they chose. Ivanantuns is right too, regarding the service content.
You use Digital Terminal Charts on your iPad or a Windows device, and NavData is loaded into FMS.


text:
Where can i find official list of all ICAO Docs published?
ICAO Docs are very essential for pilots to completely understand a concept and it is increasingly difficult to find relevant resources in internet.
It would be better to know the list of Docs available and select the docs relevant to the subject and download and study accordingly.
Hence i would like to know the official list along with frequent updates if there any amendments to docs
Thanks!

Go to icao.int -> Information Resources -> Publications -> Products & Services Catalogue (https://www.icao.int/publications/Pages/catalogue.aspx)
In this page, you'll find "Edition 20XX" (XX -Current Year) which means all the annexes and docs published before 20XX
At the bottom, you'll find "Supplements", Any Docs that has been amended or corrigendum issued in the current year-20XX will be issued as supplements
For eg. (This eg has been documented on 25.12.2022)
If you look at Page.60 of "Edition-2022" catalogue, you'll find Doc9718-Vol-2 latest edition is published in 2013
Whereas go to supplements-11 and check whether doc9718 has been amended in the current year and if you see, yes on page-2, doc9718 has been amended in 2022, so make sure you download the latest edition.
Steps to make sure you are downloading the latest edition:
1.Look in the catalogue to find the relevant doc that suits your subject under study
2.Check in the latest supplement to find any amendment/Corrigendum issued in the current year
3.Download/Purchase the latest edition of Doc.

You can find a list of International Civil Aviation Organization (ICAO) documents on the ICAO website at the following address:
https://elibrary.icao.int/explore;mainSearch=1
On this page, you will find links to all ICAO documents, organized by document type and category. You can also use the search function to find specific documents by keyword or document number.

Docs for eg. Doc-10066 are not available. I have checked this link earlier and it doesn't contain all the Docs.


text:
Why should the optimum position of GPS Antenna be close to aircraft's CG?
In CAE OAA Book11-Radio Navigation Chapter-18 Page-318, under GPS Error, "The optimum position for the antenna is on top of the fuselage close to the aircraft's centre of gravity".

In the FCOM of A320, GPS(MMR) Antenna are situated in the front of the fuselage

My questions are:

Why are all GPS Antenna situated on top of fuselage ?. (Is it because of satellite reception)
Why is the GPS (MMR) antenna situated in the front top of the fuselage ?.
Why GPS antenna should be close to aircraft's CG ?.
Why are there 2 GPS Antenna ?. (Is it because of redundancy or because faster lock-in?.)

This is pure speculation, but maybe it's because CG is basically the "centre" of the aircraft; on large aircraft, the extremities might be quite far away from the CG, and placing the antenna there might reduce the accuracy (e.g. if the antenna was placed at the wingtip, the location it would show might not give a good picture of the "average" location of the aircraft).

@AdityaSharma To avoid speculations, i have attached image from an official book. If it is given, there will be reason behind it.

Surely there must be a good reason; I was just sharing my thoughts. Good day.

Not a full answer, but consider that when a long aircraft changes pitch attitude, the nose and tail will move by several meters in opposite directions. Placing the antenna near the CG mitigates perceived altitude changes due to this effect, especially during altitude-sensitive phases of flight, like on an RNP approach.

Is it not true that an aircraft has basically, five points of reference? I deviate from standard usage to call the points front and back; left, right and centre… assuming the dheight of the body is too slight to matter
Left and right can't matter. Wingtip transponders gain what?
That leaves - untechnically - front, back and centre and takes us back to Tyzoid

The reason for "on top" is that most airplanes are flown with the top pointed more-or-less towards the sky, which is where the satellites are. If, on the other hand, one is flying their aircraft with the top pointed more-or-less towards the ground (such as while keeping up foreign relations,) then having the GPS (and other satellite) antennas on the bottom would be better. Similarly, it's better for most antennas that communicate with ground stations to be placed on the bottom of the aircraft when not flying inverted.

GPS signals need line of sight to the satellite, so positioning the antenna on top of the aircraft maximizes the satellites that can be seen and minimizes multipathing, and also gives some protection to ground based interference/jamming.

Not listed in the CAE but there are some advantages in not having critical cables running half the length of the aircraft, so the location may be the one that got a cheap, simple and easy to install and test location for the antenna, and per below it may not make a difference for this particular nav system.

Keeping the antenna near the C of G has two two effects, the first is that it means the position output is near the actual aircraft center, avoiding issues with systems having it off the side of the taxiway or showing altitude while still on the runway. There are also several effects the reduce the precision if there is rapid motion of the antenna during fixing, so a central location will get more accurate position. All of this can be 'solved' with software, but does make accurate GPS dependent on accurate pitch/roll information (and good software). Presumably in this case software was traded against the other effects of central location to produce the current placement.

Two antennas would be for redundancy. While aircraft can operate without GPS, it generally increases crew workload which produces risk of some other problem causing a crash.

To be more specific about point 3, while an aircraft is pitching, rolling, or yawing, it more or less rotates about the CofG.  So putting the antenna there keeps it moving with the path of the aircraft as a whole, not on a long level far from that. (The roll axis would be inside the fuselage, except at the nose/tail, so really it's pitch/yaw that you're fully avoiding by placing it at CofG.)

All GPS signals are line of sight and most receivers use all
satellites that are visible and a minimum of 9 - 12 degrees above
the horizon. So being on the top of the aircraft near the centerline
is essential to providing the best reception.
It's somewhat arbitrary as all antenna locations are a tradeoff. All
the various antennae have to separated to avoid potential
interference and parasitic effects. Cable lengths are an issue for
two reasons. Signal attenuation and weight. Most of the antenna
cables end up routed to the E/E bay aft of the nose gear well where
the majority of the avionics are located (HF and radio altimeters
are located in an aft bay.) I don't know why this diagram shows 4
GPS antennae, 2 MMR and 2 'GPS'. I only know of the GPS receivers in
the MMR. Every aircraft will have some variation is locations. The
biggest known interferer for GPS is the VHF com, so keeping them
separated is essential. Forward locations will avoid any shadowing
by the vertical stabilizer/rudder.
The aircraft C.G. is the reference point for the navigation system.
It's the point about which all aircraft rotation is measured. As the
nature of the GPS solution is time difference of arrival of the
signals from the satellites, the resolved position is that of the
GPS antenna. The antenna location offset from the aircraft C.G. is
loaded into each GPS(MMR) to allow the GPS to compensate for the
offset. Additionally, the GPS data will be merged with the IRS data
(which has its own compensation for installation location) in the
FMS to obtain a hybrid GPS/IRS position. So ultimately any offset
from the C.G. is accounted for. The only benefit of being close to
the is to reduce 'noise' due to aircraft rotational motion
(pitch/roll/yaw). Since it will always be present with even small
offsets, there is an internal Kalman filter to reduce it to a
minimum.
Navigation is an essential function whose loss is considered Major
or Hazardous depending on phase of flight. In either case it
requires redundancy so there are two of everything - GPS(MMR), FMS,
etc. There may be 2 or 3 IRS depending on the system architecture
and type of flight controls. Antennae can be shared between systems
if they are simple antenna and meet a minimum reliability. Because
GPS antennae are complex, containing an amplifier and downconverter,
each GPS receiver has a dedicated antenna.

Specific to question 2, that has not been addressed in other answers…  The GPS MMR antennae receive GPS augmentation signals.  In the case of WAAS, which is provided by stationary satellites, (SBAS), you want those on top.  But MMR augmentation also includes ground-based signals, (see GBAS), which are transmitted from the airport.  GBAS airports are rare today but are likely to become more common over time.  So the logical location that is line-of-sight to both satellites and the runway while on final approach is above the windshields.

The GBAS augmentation is broadcast using VHF Data Broadcast (VDB) on LOC/VOR frequencies, which the GPS antennae are incapable of receiving. Since the MMR has a perfectly functional VHF receiver (traditionally for LOC), it repurposes this hardware with supplemental software to receive the VDB signals via the LOC antenna when the GBAS landing function is selected.


text:
What is the difference between data link transmission over VHF and VDL (VHF Data Link Mode-2)?
Early FANS 1/A used ACARS network to transmit data link messages from applications like CPDLC & ADS-C using VHF or HF.
FANS 1/A+ used the same ACARS network to transmit data link messages from applications like CPDLC & ADS-C using VDL (VHF Data Link mode-2), HFDL, Satcom.
How different is datalink transmission over VHF and VDL ?


text:
How long does it take to refuel a big jumbo jet?
How long does it take to refuel a big jumbo jet? What about a smaller A320?

It's not really clear what is meant by "jumbo jet". A 777? An A340? The ICAO uses the categories heavy, medium and light for aircraft; I think that the category super (i.e. super-heavy; only the A380 is in this category) is not an official one but is still recommended for use in ATC communications. I think most jet airliners are "heavy".

The categories are for traffic separation due to wake turbulence. The category "super heavy" does exist at this time it is only used for the A380.

It depends on a number of factors like:

the aircraft size and variant
the range required
the aircraft load
equipment available
Ground crew proficiency

...among others things.
This figure from the 737 Airport Planning document gives 9 minutes for the fueling time for a Boeing 737-600:

Image from B737 Airplane characteristics for Airport Planning
The following image shows the time taken for refueling a Boeing 747-8 Jumbo to be 44 minutes:

Image from B747-8 Airplane Characteristics for Airport Planning

I need the name of the airline that can deplane in 6 minutes and board in 9. They are going to get all my business from now on. Or 12 minutes to board a 747.

It would be nice to just list the times in the chart for fueling so a reader doesn't have to figure out the chart to get the number.  Also, text is more search-friendly.

Interesting factoid: takes longer to offload crap than to load fuel in a 737....

@CGCampbell - That's hardly surprising.  Fueling is a relatively simple homogeneous process.  Essentially you just plug the hoses in and turn on the pumps until you're done.  (There's obviously a bit more to it, but it's mostly routine, predictable.)  Moving cargo involves humans making judgement calls on where to put things so they don't fall over, trying not to damage things, balancing the load, making everything fit, making sure the luggage goes where it needs to go, etc.  Much more room for variation in that sort of thing, so you have to include fudge-time.

@RonBeyer nothing is said about number of stairs/jet bridges used, neither about physical conditions of self loading fret. If you unboard a 747 full of athlets without hand luggages using the maximum number of stairs, it may be possible. In real life, it is another story.

@ManuH - I don't think I've ever seen a plane using more than one jet bridge or stairs.  I've always wondered why they don't use more - it would be much faster.  But I suppose it's a cost-saving issue.  Also, as those charts show, even if you cut the boarding time in half, there's other stuff being done at the same time that can't be expedited, so maybe there'd be little advantage to the airline to double up on the entrances.  (Would be nicer to the passengers to let them board a bit later, but what airline cares about that?)

@DarrelHoffman the question of boarding by more than one jetbridge is addressed in this question. In  short, it depends on the round trip time (low cost airliners spend less time grounded) and the size of the aircraft (a A380 has more room to put more jetbridges)

@ManuH If we only lived in an ideal world with spherical cows in a friction-less vacuum everything would be easier.

@DarrelHoffman I'm pretty sure CGCampbell meant that it takes less time to pump in the fuel than it does to "service vacuum toilets"

It depends on how many equipment are being used to refuel and how much fuel an aircraft needs. But in general there are some ratings which manufacturers issue for handling and ground time.
For example, a B747-8 has (according to 747-8 Airplane Characteristics for Airport Planning):

maximum usable fuel capacity of about 59,734 U.S. gallons or 226,113 liters
8 fueling nozzles
max fueling rate of 500 US gpm  (1,890 lpm) per nozzle
total max fuel pressure 50 PSIG

Therefore, in theory it's possible to refuel an empty B747-8 to its maximum usable fuel capacity in 15 min with all 8 nozzles. In normal refueling configuration two trucks (one truck each wing) being used which serves 4 pressure nozzle and takes 30 min to refuel 56,553 US gallons. So standard fueling time for this aircraft estimated about 50 min.
For A320-200 though these ratings are (according to Airbus A320 airplane characteristics for airport planning):

maximum usable fuel capacity of 6,303 U.S. gallons or 23,859 liters
6 fueling nozzles
max fueling rate 369.84 US gpm  (1,400 lpm) per nozzle
total max fuel pressure 50 PSIG

And the estimated fueling time is about 15 min refueling 6,303 U.S. gallons using one truck serving two nozzles.

This answer seems to be saying that an A320 has only one tenth the fuel capacity of a 747.

@ChrisW Yes, that's what it's saying. It's true.

@ChrisW - This may clear things up a little :) a 747's engines are nearly the same diameter as an A320 fuselage... https://c2.staticflickr.com/6/5528/14609549344_1d5c8f9d3e_b.jpg

@ChrisW yes. Based on those official documents and other technical documents. Another example is 737ng with usable fuel capacity of 20,800kg.

@JonStory And even the 747's engines look small when mounted next to a GE90-115b (the engine for the 777-300ER.)

I worked for a long haul airline with 744s departing South East Asia to Europe.
The airplane usually required 120-140 metric tonnes of fuel for the trip back. There would usually be around 15-20 tonnes left from the previous sector and you would usually require around 30-40 minutes to fuel 100-120 tonnes. This is using two pumps, one on each wing.
The problem was if you needed to fuel up to max which was around 170 tonnes (depending on the fuel density), the last 10-20 tonnes took longer to fuel. This was rare though. Since we only had  slightly more than an hours ground time we had to board while fueling was in progress. There is an SOP for this, basically a door had to be open at the rear and a set of steps positioned, we usually used door 5L. Also passengers were told not to fasten seat belts, crews were required to be at their stations and most importantly a firetruck had to be standing-by at the stand. We had to pay for the firetruck call-out but on-time departure was more important.

Thanks for this interesting first-hand account! Out of curiosity, why did the last tonnes take longer to fuel?

According to this article, it generally takes about 1-2 hours to fuel a large jumbo such as an A380. As for smaller jets, the second answer to the Yahoo! question says that Southwest Airlines claims to be able to turn a plane around in 20 minutes. Given that the Southwest fleet is made up entirely of 737s, a jet such as the 737 or A320 should take about 15-20 minutes to refuel.

I don't think it's out of the realm of possibility that quick turn around flights carry enough fuel for multiple legs. Usually the quick turn arounds are short regional hops and the airline may purchase fuel where it's cheaper.

@RonBeyer most airlines do not buy fuel like you and I do at the pump. See fuel hedging. Chances are, they will fill up when required.

@RonBeyer The more fuel you have loaded, the more fuel you will burn just to carry it around. Ryanair has been fined because they repeatedly carried so little fuel that they fell below minimum required reserve during the first or second hold pattern, so the captain had to declare an emergency. If Ryanair acts like that, it must save them a buck or two...

@BurhanKhalid, purchasing and hedging are two separate activities. Or at least often are. Goldman Sachs enters into and/or brokers hedging contracts. No one at Goldman Sachs is pumping fuel into a 737. Nor is Goldman Sachs contracting with anyone to pump fuel.


text:
Why do pilots use airways instead of just "flying direct" every time?
These days, on larger/newer aircraft at least, GPS navigation is ubiquitous.  So VORs and Non-directional Beacons (NDBs) are used less.  Wasn't the reason for airways because they were the connection points between VORs and NDBs?
It seems like a pilot could easily "fly direct" from one airport to another without using any intermediary checkpoints.  And yet I've been told on several occasions that most pilots use airways.
Why do pilots still use airways instead of flying direct?

I can't seem to find any references at the moment, but I distinctly remember an Air Crash Investigation episode that stated that flying direct was the norm in the early days of aviation; until a spate of accidents forced the FAA to establish airways to control this.

Interesting, if you can find it @DannyBeckett, I'd be curious to see it.

Some areas with a lesser density of airports, mostly in Scandinavia, use direct routing and no airways from certain fixes onward. http://www.eurocontrol.int/sites/default/files/article/files/2011march-free-route-airspace-maastricht.pdf

@DannyBeckett That may well be true but the original reasons for establishing airways might no longer apply. For example, if it was just to make navigation easier, major airlines would probably turn around and say, "Thanks but we have GPS and computers and stuff. We don't need that help."

On a flight from JAC-MCO (~1700nm) in a PC-12 (NBAA IFR Range 1600nm), I filed a route on the backside of a low which was hovering over the direct route. Since ZLC is so sparse, every new controller gave us "Cleared direct MCO," and I kept telling them we want "as filed." We made MCO with plenty of fuel to spare, and it was the longest endurance flight we ever did (yes, it had an onboard potty).

@Jay Carr...I think the episode in question dealt with a collision between two airliners over the Grand Canyon in the '50s(?).  Both flights deviated to give passengers a view..

Airways simply allow for better management of traffic.
Imagine for a moment that everyone had an off-road capable car, if all the drivers were going "GPS Direct" to their destination how would drivers ensure separation? How would you avoid hitting other cars if there were no roads?
Airways are the aviation solution to this problem: Defined routes between navaids (VORs, NDBs, and the like). Airways can also be defined using "fixes" defined by GPS coordinates which can provide something closer to "GPS Direct" routing while still providing a defined route.
Airways also provide a bunch of other benefits, including:
Traffic Flow Management
Flightplans in Europe are validated by the CFMU (Central Flow Management Unit) and checked for criteria such as direction of airways (which can change during the day, e.g. one airway being for eastbound traffic only between 1000 UTC and 1400 UTC), correct flight level ranges and airways correctly connecting waypoints, such as fixes, VORs or NDBs. The bottleneck in all flight operations is understanding airport capacity (how many operations/hour) and sector capacity (how many aircraft handled per sector/per controller). The goal is sometimes to distribute traffic between two sectors, although having the same destination. Below you will find two routings to Düsseldorf - EDDL, one departing from Leipzig - EDDP and one from Berlin-Tegel - EDDT (click on links to see visual representation):
EDDP - EDDL
ORTAG Q230 WRB T854 DOMUX

EDDT - EDDL
BRANE Y200 HLZ T851 XAMOD

The routing from EDDP to EDDL is going via WRB - TINSA - ADEMI - INBAX - DOMUX, which is within the boundaries of a sector called "Paderborn High" or PADH.
The routing from EDDT to EDDL is going via PIROL - DENOT - HMM - XAMOD, which is within the boundaries of a sector called "Hamm High/Medium" or HMMH/HMMM.
During peak hours, both sectors are separate and deal with traffic only within their sector, which uses the full capacity of each controller handling the sector. Outside peak hours, the sectors can be combined and worked by one controller. This allows a granular usage of controller resources and airspace.
Separation between Traffic
From the example above we have seen that there is 2 routing endpoints at Düsseldorf - EDDL for traffic arriving from the North-East and South-East, XAMOD and DOMUX. Both waypoints or fixes are respective STAR or transition entry-points for arrivals into Düsseldorf. By using a standardized routing to and from airports, we can now expect traffic to always arrive via these two points from the east, unless other coordination has been achieved by the ATC units involved. Looking at the reverse routing for both airports, we will see these routes:
EDDL - EDDP
NUDGO Z858 BERDI Z21 BIRKA T233 LUKOP

EDDL - EDDT
MEVEL L179 OSN L980 DLE T207 BATEL

MEVEL and NUDGO are two of the SID exit waypoints out of Düsseldorf, MEVEL being 10nm north of XAMOD and NUDGO being 26nm east of DOMUX, however the relevant fix here is ELBAL, which is one of the fixes on the SID23L from Düsseldorf to NUDGO, which is 16nm south of NUDGO. With something simple as using standardized entry and exit fixes to and from an airport, we have managed to maintain a traffic flow with ensured separation between departing and arriving traffic, something which would not be possible with GPS direct routing or would require constant ATC vectors.
Handoffs from ATC Unit to ATC Unit
The above examples show different routings and sometimes different sectors handling traffic on these routes. How is traffic handed off to other controllers and where? A handoff consists of moving the radar track of the aircraft from one controller to another and thereafter instructing the aircraft to change the frequency to the next sector controller. Handoffs are initiated at sector boundaries, at specific fixes, which have been agreed on between different ACC (Area Control Centers) or sometimes even within one ACC between single sectors. Since we are already familiar with Düsseldorf airspace, let's use the below routing from München to Düsseldorf.
EDDM - EDDL
GIVMI Y101 TEKTU Z850 ADEMI T854 DOMUX

From the graphical represenation you will see that traffic from EDDM to EDDL will pass the waypoint ARPEG, which is close to the destination airport. Before reaching the waypoint ARPEG, aircraft will be under control of either the Hersfeld (HEF) or Gedern (GED) sector (page 5) and need to descend towards the destination airport. The next sector for arriving traffic into Düsseldorf via ARPEG would be Paderborn High Sector (PADH). The agreements between HEF/GED and PADH are that arriving traffic to Düsseldorf via ARPEG is to be expected at FL240 level at ARPEG and released after handoff for further descent through PADH, even if the aircraft has not yet crossed the sector boundary and is not in PADH airspace. These coordination waypoints and handoff procedures are documented in Letters of Agreement between or within an ACC.
Weather
When weather (wrong-way lows, strong jetstreams, storms, etc) affects the Direct route, flight dispatchers will pro-actively suggest routes around it. It's not unusual for NYC-SFO traffic to be routed over Canada or over Kentucky when the Midwest is storming.
Exceptions - Free Route Airspace
Where airport and air traffic density allows, the use of airways is not mandatory or in some cases, not even expected or provisioned for. One example is the Free Route Airspace overhead Scandinavia, where a portion of airspace can be freely used between defined entry and exit points. Similar systems are also being discussed by NATS UK and the  FAA NextGen Air Transportation System.

For more information on airways and their use, see also the related question:
Is there a difference between how commercial jets and GA aircraft use airways?

Some material courtesy of VATSIM or VATSIM Germany. While only for simulation use, the material used is as accurate and close to the real sectorisation or procedures as possible, so should be sufficient to explain the concepts presented here.

I'll try to create and show examples tomorrow when I am not commenting from a mobile device. :)

Excellent car analogy

Roads aren't there for traffic control; they're there because cars are inefficient and (even somewhat more than usually) unsafe off-road, with or without any other cars around. This analogy seems highly dubious.

@JayCarr I have provided more examples for you.

@NathanTuggy I have amended my analogy, I had the assumption in mind that all cars were off-road capable, same as all aircraft are flight-capable. Otherwise the analogy wouldn't make sense, as you have pointed out.

The Say Again series on AvWeb discusses using the routes many times over. My understanding from it is that ad-hoc routes are not big problem in radar environment, but nothing but departure and destination airports is. At the very least the flight plan should have departure point (wehere SIDs end), initial approach fix (where STARs start) to destination and one point after each sector transition so the controller will have a known reference point for your track and point to hand you over at.

@JanHudec The Say Again series are dominantly FAA-based, so I can't comment. In Europe, you will rarely see non-CFMU validated flightplans due to the complexity and density of the airspace.

@SentryRaven: The Say Again series also talk mainly about low-flying general aviation aircraft and that they often don't bother creating proper route and the problems it creates, while airlines generally do file proper routes. But then they usually have dedicated people in dispatch for it.

@JanHudec Your comments made me so curious that I devoted another question to what you're talking about.  If you want to expand your thoughts there I'd love to hear them:  http://aviation.stackexchange.com/questions/11514/is-there-a-difference-between-how-commercial-jets-and-ga-aircraft-use-airways

Flying specific airways and arrivals also makes separation in non radar environments possible and more predictable.  It also helps on lost comm situations since anyone looking at the flight plan knows when the aircraft will commence.

One additional point relevant to light GA aircraft and off-airway routing is separation from terrain on an IFR flight plan. On airways in difficult terrain, the airway is checked for obstacle clearance above the MOCA, and navaid performance and communications are checked at the MRA, and aircraft are normally routed at or above the MEA. Off-airway, obstacle avoidance is only guaranteed above the OROCA, which does not guarantee navaid reception, communications, or radar coverage. (SIDs/STARs serve similar functions plus help traffic flow in terminal environments.) This is FAA-specific, probably.

A simplistic explanation is,
pilots in controlled airspace fly along airways because that's where
the air traffic controllers want them to fly.
The use of airways has to do with the way air traffic controllers keep track of all the
flights in the sector of airspace they control
(they need a way to organize the traffic so that they have a good mental
image of it and can guarantee that aircraft do not pass too close to each other),
as well as the fact that it is easier to communicate a path that
goes between published waypoints with three- or five-letter names than
a path that goes between points described by their numeric latitudes and longitudes.
The free flight concept
proposes to allow these same aircraft to fly routes that do not
generally follow airways, but it requires big changes in the way
air traffic management is done.
(This answer is based primarily on recollection of the
work I've done on the development and analysis of air traffic management
systems, mostly related to free flight, since 2001.)

"because that's where the air traffic controllers want them to fly" -- I think this is not true historically. Many airways were developed because of terrain, radio coverage, and city-pair traffic.

@rbp I did say "simplistic". The point of the OP, however, which I think was valid, was that the available choices of flight paths are much more limited (most of the time) than current navigation technology could support. The reasons I usually hear cited for this basically come down to safety, specifically, maintaining separation between multiple aircraft, which is a very different problem than the problem of navigating an individual aircraft.

@rbp P.S.--Loved the story about your JAC-MCO flight.

fair enough. Glad you liked the story

Airways are safer, because you know where other pilots will be flying in order to avoid them. The rule of using an odd flight level for flying east, and an even flight level for flying west will also help to keep traffic separated.
Safe cruising altitudes are also set based on the airways. There is an altitude per waypoint so a normal climb will let you get over the next hill, and per section between waypoints so you know how high that hill actually is.
Without the airways system maintaining proper ground clearance would be much more work intensive. With the fixes all the pilot needs to do is keep a list with fixes and MEA/MCA (safe crossing altitudes) and follow the needle and cross them off as he reaches them and climb if necessary.

In the absence of separation between eastbound and westbound flights, would collisions be more likely between eastbound and westbound flights, or between flights going in roughly the same direction? I would think the dangerous scenario from a collision standpoint would be two aircraft flying almost parallel such that neither one ever entered the field of view of the other.

Just to add a little perspective, route-free airspace has been in consideration for a while and is already implemented in certain countries. The route-free or direct-routing concept makes use of the currently underutilized satellite systems on aircraft (In the routing context) and intelligent software to execute more efficient routing of aircraft (GPS-based). Direct-routing is also a part of the the NATS UK and the FAA's NextGen air traffic management strategy.

Speaking as a GA pilot flying light airplanes in the US system:
On flights in lower use airspace, at lower altitudes, and with the appropriate ATC controls in place (agreements allowing controllers to do that across into someone else's airspace) it's possible to get a filed GPS direct route approved.  I've done so on trips in the 200-300 mile range before.
On longer trips, you don't necessarily have to use the airways.  It's often possible to get waypoints at the ingress/egress of airspace, which you can go direct between (off airway).

It's like a lot of systems that evolve.  If we started today, we might do it differently.
The analogy with cars is a poor one because, for the most part, there are a very small number of aircraft in a very large amount of airspace and it would be hard to justify the use of airways.  So starting from the standpoint that every aircraft accurately knows:
a) Its location, heading, velocity vectors, fuel data, flight constraints and destination etc;
b) The same information for all other aircraft in its vicinity (this could cover quite a large volume in practice);
c) The terrain and weather conditions etc.
Then it becomes a really interesting combinational problem to solve in the context of an autonomous distributed computer system; i.e. no ATC - all handled by cooperating in-flight systems.
Clearly there are significant points where paths must cross because of the geographical positions of airports, and of course the dynamic scheduling of arrivals and departures at airports; not to mention the handling of unexpected events.
For a computer scientist this is all fascinating and a meaty challenge, but I suspect that it would be very difficult to develop a system of autonomous control based on this approach to the point that it could be trusted!

What statistical data do you base the assumption on that there is a small number of aircraft in a very large amount of airspace? Especially around busy arrival airspaces, this is quite not true.

The question was about the need for airways in general, not arrival airspaces.  Even taking the busiest time for the busiest country in the world, aircraft (with a reasonable separation) currently occupy only a tiny fraction of the available space.  Even close to a busy airport with up 200 flights/hour, a basic cubic law indicates that any congestion need only be restricted to a very short distance around it.

Not sure I agree with "a" completely. A light GA (without GPS) aircraft knows it's general location, but to what numerical value? Don't properly adjust your altimeter for change in weather and maybe you're off by a few ten's of feet. There ARE still aircraft out there flown by pilots without GPS, without even a radio. I also totally disagree with "b". My father, flying his old Piper, has no idea of all other aircraft in his vicinity. He knows approximately the locations of those he can see visually, but that is it.

I was making the assumption that at some point it will be automatic and mandatory for all aircraft, much as it will be for cars on highways in the near future.


text:
Why do we need to enter squawk code when mode-s has ICAO 24-bit address?
Mode-S transponder was introduced to to eliminate reduced codes available to designate aircraft (4096 squawk codes) in a congested airspace.
With the introduction of mode-S transponders, unique ICAO 24-bit address (17million codes) was permanently assigned to an aircraft.
But still squawk codes are assigned to mode-S transponders, why ?
Example of ICAO 24-bit address: (taken from flightradar24.com), 89611E

Mode S was not introduced because of a shortage of Mode A (squawk) codes in congested airspace, but to improve the radar performance.
In congested airspace, a Mode A/C radar suffers from two technical phenomena: garbling and FRUIT.
Garbling occurs when the reply of two transponders overlap, causing the radar to have trouble in distinguishing the replies. This happens when aircraft are nearby each other.
FRUIT (Foreign Replies Unsynchronised in Time) occurs when a radar interprets a transponder's reply to an interrogation by another radar as a reply to its own interrogation.
Mode S (S = Selective) addressed both of these problems by introducing addressed interrogation, allowing radars to selectively solicit for replies of aircraft. For this the ICAO 24-bit address is used, uniquely assigned to each aircraft. Radars have a similar, albeit non-unique address, the IC/II (interrogator code / interrogator identifier).

So why can't the 24-bit address replace the 12 bit squawk code?

not everywhere on earth Mode S radars are used. There are still Mode A/C only radars around. For a long time, it was very common to have Mode A/C only. For the same reason, not every aircraft is equipped with a Mode-S transponder.

the squawk codes have a number of functions, some cannot be replaced by a static 24-bit address.

correlating the radar track to a flight plan. This could be done based on 24-bit address
signifying the nature of the flight (e.g. VFR 7000/1200 codes, 2000 code, dedicated codes for Helicopter Emergency Service etc) - this cannot be done based on the 24-bit address
indicating emergency status (7500, 7600, 7700), this cannot be done based on the 24-bit address either.

In order to address the shortage of squawk codes, one of the approaches taken is to use the Flight ID (call sign) transmitted by the Mode-S transponder in lieu of the Squawk code for associating the radar track to a flight plan. In this case, the squawk code 1000 is used to signal to the flight data processing system that association shall be done base on Flight ID.

"indicating emergency status" Mode-S has built-in facilities for doing this, see Type Code 28/BDS 6,1, which encodes more emergency status, including lifeguard/medical and minimum fuel. Mode-S transponders current automatically output this message depending on Mode A setting.

@user71659 TC 28 / BDS 6,1 is part of 1090ES (ADS-B), not of the Mode S radar protocol. But it is a fair point, and shows that you know an awful lot about the nitty gritty details of ATC surveiilance.

In some countries, if the equipment codes in the flight plan indicate mode S capability, the computer will assign a squawk code of 1000, which is a special code that causes the transponder to not transmit mode A at all.  This works because in those countries, ATC’s software can use the mode S aircraft ID (aka “hex code”) to correlate a target with a flight plan.
However, not all countries have upgraded their software to use the mode S code this way, so they still need the mode A code for that purpose.

Excellent explanation. But this led me to few more questions.

If the equipment on board an aircraft is mode-S transponder and still ATC asks aircraft to squawk certain 4-digit code, how will the transponder behave now ?
Will it transmit mode-S data (24-bit address along with FL, Speed, Callsign, Lat & Long) and mode-A code or upon entering 4-digit code, the transponder downgrades its capability and transmits only mode-A/C information (Squawk code + altitude) ?

@StephenS do you have any examples of ATC systems doing correlation on 24-bit address when Mode =1000? Normally the correlation is done based on FlightId/Callsign for Mode A = 1000 tracks.

@KrishnaKanth Mode S transponders also send mode A/C information, except mode A is suppressed if the squawk code entered is 1000.


text:
What is the difference between RNAV-Terminal usage waypoints vs RNAV-STAR waypoints?
What is the difference between [MM604 (or) MM615] vs [RADAB (or) DAMAD] in the STAR attached below

When seen using Enroute Chart
The description for RADAB or DAMAD is:

The description for MM604 or MM615 is:

What does does this 2 RNAV waypoint have a difference and how it is well interpreted in Airbus A320 FMGS F-PLN page in MCDU ?.

What "difference" are you seeing that you're asking about? I'm not seeing anything different between the two screenshots of the details beyond what's already obvious by considering the waypoint names (i.e. pronounceable vs MM###). What difference are you drawing attention to with the screenshots?

@RalphJ There is regulation on how enroute waypoints should be named (2 vowels and 5 alphabets). Similarly is there any naming convention followed to name star waypoint ? I am sure the difference must be more than just place of waypoint( enroute vs STAR/SID). kindly provide regulation document if you have it for SID/STAR Waypoint naming. Thanks